The fraction will be 15/36.
Depending on the result of the second roll, break it down by cases:
1: {null} = no result
2: {1} = (2,1) = 1 result
3: {1,2} = (3,1), (3,2) = 2 result
4: {1,2,3} = (4,1), (4,2), (4,3) = 3 result
5: {1,2,3,4} = (5,1), (5,2), (5,3), (5,4) = 4 result
6: {1,2,3,4,5} = (6,1), (6,2), (6,3), (6,4), (6,5) = 5 result
So, total 15 out of 36 total pairs fit the criterion = 15/36
That is the likelihood that, if a die is rolled twice, the second roll will be less than the first.
So, our final answer will be 15/36.
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